Solve x^3-1/64

$$x ^ { 3 } - \frac { 1 } { 64 }$$

Factor

$$\frac{\left(4x-1\right)\left(16x^{2}+4x+1\right)}{64}$$

Solution Steps

Factor out \frac{1}{64}.

$$\frac{64x^{3}-1}{64}$$

Consider 64x^{3}-1. Rewrite 64x^{3}-1 as \left(4x\right)^{3}-1^{3}. The difference of cubes can be factored using the rule: a^{3}-b^{3}=\left(a-b\right)\left(a^{2}+ab+b^{2}\right).

$$\left(4x-1\right)\left(16x^{2}+4x+1\right)$$

Rewrite the complete factored expression. Polynomial 16x^{2}+4x+1 is not factored since it does not have any rational roots.

$$\frac{\left(4x-1\right)\left(16x^{2}+4x+1\right)}{64}$$

Evaluate

$$x^{3}-\frac{1}{64}$$